Write 1 and 1/5 and 1 and 1/3 using a common denominator :)

The following scatter plot represents the relationship between a person's weight and the number of calories the person burns in

Kamila [148]

Answer:

Positive correlation.

Step-by-step explanation:

This is a positive correlation. Because the dots don't jump around too much, they are steadily moving in one direction, so they have correlation.

But also, the dots are moving upwards, in a positive direction. So, that's why they have a positive correlation.

3 0

4 years ago

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Find an equation for the tangent to the curve at P and the horizontal tangent to the curve at Q. y = 5 + cot x - 2 csc x 0 1 2 3

oksian1 [2.3K]

Answer with explanation:

The given function in x and y is,

y= 5 +cot x-2 Cosec x

To find the equation of tangent, we will differentiate the function with respect to x

$y'= -\csc^2 x+2 \csc x\times \cot x$

Slope of tangent at (π/2,3)

$y'_{(\frac{\pi}{2},3)}= -\csc^2\frac{\pi}{2} +2 \csc \frac{\pi}{2}\times \cot \frac{\pi}{2}\\\\=-1+2\times 1 \times 0\\\\= -1$

Equation of tangent passing through (π/2,3) can be obtained by

$\rightarrow \frac{y-y_{1}}{x-x_{1}}=m(\text{Slope})\\\\ \rightarrow \frac{y-3}{x-\frac{\pi}{2}}=-1\\\\\rightarrow 3-y=x-\frac{\pi}{2}\\\\\rightarrow x+y-3-\frac{\pi}{2}=0$

⇒There will be no Horizontal tangent from the point (π/2,3).

5 0

4 years ago

A major credit card company is investigating whether the distribution of the number of credit cards used by its customers has ch

Mumz [18]

Answer:

C) If the null hypothesis were true, there is a 2 percent chance of obtaining a chi-square value of at least 7.82.

Step-by-step explanation:

The value of the chi-square test statistic was 7.82 and the p- value is 0.02. This means that if the null hypothesis were true there is a 2 percent chance of obtaining a chi square value of at least 7.82

The p- value is defined as obtaining the test statistic under H0 as extreme as observed.

When the p- value is less than 0.05 in the chi square goodness fit test it means that the null hypothesis is rejected.

There fore option C is the best answer.

A)There is a 2 percent chance that the company's claim is correct.

B) There is a 2 percent chance of obtaining a chi-square value of at least 7.82.

C) If the null hypothesis were true, there is a 2 percent chance of obtaining a chi-square value of at least 7.82.

D) If the null hypothesis were true, there is a 2 percent chance that the company's claim is correct.

E) If the null hypothesis were true, there is a 2 percent chance of obtaining a chi-square value of 7.82.

5 0

3 years ago

Help me look at the picture

dexar [7]

Answer:

it's B

Step-by-step explanation:

lol how do you not know this

8 0

3 years ago

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Qualification exams for becoming a state-certified welding inspector are based on multiple-choice tests. As in any multiple-choi

erik [133]

Answer:

(a) The probability distribution of the random variable X is Binomial.

(b) The mean and standard deviation of the 25-question test are 5 and 2 respectively.

(c) The value of x is 0.34.

Step-by-step explanation:

The random variable X is defined as the number of correct answers given by a person who is guessing each answer on a 25-question exam.

There are five possible answer for every question.

This implies that the probability of getting a correct answer is:

P (X) = 0.20.

There are a total of n = 25 questions.

Every answer is independent of the others.

(a)

The random variable X has finite number of independent trials (i.e. 25 questions). There are only two outcomes for each trial, i.e. Success = correct answer and Failure = wrong answer. Each trial has the same probability of success (, i.e. P (X) = 0.25).

Thus, the probability distribution of the random variable X is Binomial with parameters n = 25 and p = 0.20.

(b)

Compute the mean of the random variable X as follows:

$E(X)=np\\=25\times 0.20\\=5$

Compute the standard deviation of the random variable X as follows:

$SD(X)=\sqrt{np(1-p)}\\=\sqrt{25\times 0.20\times (1-0.20)}\\=2$

Thus, the mean and standard deviation of the 25-question test are 5 and 2 respectively.

(c)

The sample is large and the probability of success is close to 0.50.

So a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

1. np ≥ 5

2. n(1 - p) ≥ 5

Check the conditions as follows:

$np=25\times 0.20=5=5\\\\n(1-p)=25\times (1-0.20)=20>5$

Thus, a Normal approximation to binomial can be applied.

So, $X\sim N(5, 2^{2})$

It is provided that the minimum passing score foe the test is such that only 1% of students who are guessing will pass the test.

That is P (X < x) = 0.01.

⇒ P (Z < z) = 0.01

The value of z is -2.33.

Compute the value of x as follows:

$z=\frac{x-\mu}{\sigma}\\\\-2.33=\frac{x-5}{2}\\\\x=5-(2.33\times 2)\\\\x=0.34$

Thus, the value of x is 0.34.

4 0

4 years ago